If in a triangle overrightarrow{AB} = vec{a} | vedclass

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(D) Given that $D$ is the mid-point of $AB$,so $\overrightarrow{AD} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}\vec{a}$.Given that $E$ is the mid-p

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$\frac{\vec{b}}{2} - \frac{\vec{a}}{2}$

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(D) Given that $D$ is the mid-point of $AB$,so $\overrightarrow{AD} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}\vec{a}$.Given that $E$ is the mid-point of $AC$,so $\overrightarrow{AE} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}\vec{b}$.Using the triangle law of vector addition in $	riangle ADE$,we have $\overrightarrow{AD} + \overrightarrow{DE} = \overrightarrow{AE}$.Therefore,$\overrightarrow{DE} = \overrightarrow{AE} - \overrightarrow{AD}$.Substituting the values,$\overrightarrow{DE} = \frac{1}{2}\vec{b} - \frac{1}{2}\vec{a} = \frac{\vec{b}}{2} - \frac{\vec{a}}{2}$.

If in a triangle $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AC} = \vec{b}$,and $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively,then $\overrightarrow{DE}$ is equal to:

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